I have recently begun to study quasi-triangular structures and have come across a problem I can't resolve. Let ${\cal U}_q({\mathfrak sl}_N)$ denote the quantised enveloping algebra of ${\mathfrak sl}_N$, and let $R$ be a universal R-matrix for ${\cal U}_q({\mathfrak sl}_N)$ . If we denote the usual dual pairing of ${\cal U}_q({\mathfrak sl_N})$ with
$SU_q(N)$ by $\langle \cdot , \cdot \rangle$, then it is well known that $$R^{ir}_{js} = \langle R, u^i_j \otimes u^r_s \rangle = q^{\delta_{ir}}\delta_{ij}\delta_{rs} + (q-q^{-1})\theta (i-r)\delta_{is}\delta_{jr}.$$ A natural question to ask is whether such a formula exists for $(R^{-1})_{js}^{ir}=\langle R^{-1}, u^i_j \otimes u^r_s \rangle$. An obvious guess would be to take the inverse of the matrix $[R_{js}^{ir}]_{i,r,j,s}$.

That is, to guess that $$[(R^{-1})_{js}^{ir}] _{i,r,j,s}$$

is equal to $$([R_{js}^{ir}]_{i,r,j,s})^{-1}.$$ This guess is confirmed by the fact that

According to Klimyk and Schmudgen's book, your formula for the $R$-matrix is incorrect. They have $\theta(i-r)$ where you have $\theta(r-s)$. That term is nonzero only when $s=i$, so your formula is the same as $\theta(r-i)$. I don't know if this anyway gives the same result, but that's all I can come up with.
–
MTSMay 22 '10 at 19:46

Thanks for pointing that out, it's now correct. Unfortunately it was just a typo with no affect on the calculations.
–
Abtan MassiniMay 22 '10 at 20:26

1 Answer
1

I think the only issue here is a harmless error in your calculation and that there is a normalization of the $R$-matrix for $U_q(sl_N)$ by a factor of $q^{1/2}$ which you have omitted (See 8.4.2 of Klymik Schmudgen).

First, I get $(R^{-1})^{21}_{12} = -q^{-1} R^{21}_{12}$,
because $\langle(S\otimes id)(R),a^2_1\otimes a^1_2\rangle = \langle R,S(a^2_1)\otimes a^1_2 \rangle = -q^{-1} \langle R,a^2_1\otimes a^1_2\rangle$, using that $S(b)=-q^{-1} b$ from Proposition 4.1.2.3 of Klymik Schmudgen. So the actual matrix you should get should just be $q^{-1}$ times what you had expected to get.

Now the factor of q^{-1} here is because you had multiplied the actual universal R matrix by $q^{1/2}$ and so $(\lambda A)^{-1} = \lambda^{-1} A^{-1}$, so there's a factor of $\lambda^2$ as a discrepancy.